Solveeit Logo
Login

Question

Question: $\lim_{x \to 0} \frac{(\cos x)^{\csc^2x}-e^{\csc^2x(\cos x -1)}}{x^2}$...

lim⁡x→0(cos⁡x)csc⁡2x−ecsc⁡2x(cos⁡x−1)x2\lim_{x \to 0} \frac{(\cos x)^{\csc^2x}-e^{\csc^2x(\cos x -1)}}{x^2}

Answer

−18e-\frac{1}{8\sqrt{e}}

Explanation

Solution

The given limit is lim⁡x→0(cos⁡x)csc⁡2x−ecsc⁡2x(cos⁡x−1)x2\lim_{x \to 0} \frac{(\cos x)^{\csc^2x}-e^{\csc^2x(\cos x -1)}}{x^2}. This is of the indeterminate form 00\frac{0}{0}. We will use Maclaurin series expansions for the functions involved.

We need the series expansions for cos⁡x\cos x, sin⁡x\sin x, eue^u, and ln⁡(1+u)\ln(1+u) around x=0x=0. cos⁡x=1−x22!+x44!−O(x6)=1−x22+x424−O(x6)\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6) sin⁡x=x−x33!+O(x5)=x−x36+O(x5)\sin x = x - \frac{x^3}{3!} + O(x^5) = x - \frac{x^3}{6} + O(x^5) eu=1+u+u22!+O(u3)e^u = 1 + u + \frac{u^2}{2!} + O(u^3) ln⁡(1+u)=u−u22+u33−O(u4)\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - O(u^4)

First, let's find the expansion for csc⁡2x\csc^2 x: sin⁡2x=(x−x36+O(x5))2=x2(1−x26+O(x4))2=x2(1−2x26+O(x4))=x2(1−x23+O(x4))\sin^2 x = (x - \frac{x^3}{6} + O(x^5))^2 = x^2(1 - \frac{x^2}{6} + O(x^4))^2 = x^2(1 - \frac{2x^2}{6} + O(x^4)) = x^2(1 - \frac{x^2}{3} + O(x^4)) csc⁡2x=1sin⁡2x=1x2(1−x23+O(x4))=1x2(1−x23+O(x4))−1\csc^2 x = \frac{1}{\sin^2 x} = \frac{1}{x^2(1 - \frac{x^2}{3} + O(x^4))} = \frac{1}{x^2}(1 - \frac{x^2}{3} + O(x^4))^{-1} Using (1−u)−1=1+u+u2+…(1-u)^{-1} = 1+u+u^2+\dots, we get: csc⁡2x=1x2(1+x23+O(x4))\csc^2 x = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4))

Now, let's analyze the exponent term that appears in both parts of the numerator: E=csc⁡2x(cos⁡x−1)E = \csc^2x(\cos x - 1). cos⁡x−1=−x22+x424−O(x6)\cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - O(x^6) E=1x2(1+x23+O(x4))(−x22+x424−O(x6))E = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4)) \left( -\frac{x^2}{2} + \frac{x^4}{24} - O(x^6) \right) E=(1+x23+O(x4))(−12+x224−O(x4))E = (1 + \frac{x^2}{3} + O(x^4)) \left( -\frac{1}{2} + \frac{x^2}{24} - O(x^4) \right) E=−12+x224−x26+O(x4)E = -\frac{1}{2} + \frac{x^2}{24} - \frac{x^2}{6} + O(x^4) E=−12+x2(124−424)+O(x4)E = -\frac{1}{2} + x^2 \left( \frac{1}{24} - \frac{4}{24} \right) + O(x^4) E=−12−3x224+O(x4)=−12−x28+O(x4)E = -\frac{1}{2} - \frac{3x^2}{24} + O(x^4) = -\frac{1}{2} - \frac{x^2}{8} + O(x^4)

Let's evaluate the second term in the numerator, B=ecsc⁡2x(cos⁡x−1)=eEB = e^{\csc^2x(\cos x -1)} = e^E: B=e−1/2−x2/8+O(x4)=e−1/2e−x2/8+O(x4)B = e^{-1/2 - x^2/8 + O(x^4)} = e^{-1/2} e^{-x^2/8 + O(x^4)} Using eu=1+u+O(u2)e^u = 1+u+O(u^2): B=e−1/2(1+(−x28+O(x4))+O(x4))B = e^{-1/2} \left( 1 + \left(-\frac{x^2}{8} + O(x^4)\right) + O(x^4) \right) B=e−1/2(1−x28+O(x4))B = e^{-1/2} \left( 1 - \frac{x^2}{8} + O(x^4) \right)

Now, let's evaluate the first term in the numerator, A=(cos⁡x)csc⁡2xA = (\cos x)^{\csc^2x}. We can write A=ecsc⁡2xln⁡(cos⁡x)A = e^{\csc^2x \ln(\cos x)}. Let F=csc⁡2xln⁡(cos⁡x)F = \csc^2x \ln(\cos x). First, expand ln⁡(cos⁡x)\ln(\cos x): ln⁡(cos⁡x)=ln⁡(1+(cos⁡x−1))\ln(\cos x) = \ln(1 + (\cos x - 1)). Let u=cos⁡x−1=−x22+x424−O(x6)u = \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - O(x^6). Using ln⁡(1+u)=u−u22+O(u3)\ln(1+u) = u - \frac{u^2}{2} + O(u^3): ln⁡(cos⁡x)=(−x22+x424)−12(−x22)2+O(x6)\ln(\cos x) = \left(-\frac{x^2}{2} + \frac{x^4}{24}\right) - \frac{1}{2}\left(-\frac{x^2}{2}\right)^2 + O(x^6) ln⁡(cos⁡x)=−x22+x424−x48+O(x6)\ln(\cos x) = -\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^4}{8} + O(x^6) ln⁡(cos⁡x)=−x22+x4(124−324)+O(x6)\ln(\cos x) = -\frac{x^2}{2} + x^4\left(\frac{1}{24} - \frac{3}{24}\right) + O(x^6) ln⁡(cos⁡x)=−x22−2x424+O(x6)=−x22−x412+O(x6)\ln(\cos x) = -\frac{x^2}{2} - \frac{2x^4}{24} + O(x^6) = -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6)

Now, substitute this into FF: F=1x2(1+x23+O(x4))(−x22−x412+O(x6))F = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4)) \left( -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6) \right) F=(1+x23+O(x4))(−12−x212+O(x4))F = (1 + \frac{x^2}{3} + O(x^4)) \left( -\frac{1}{2} - \frac{x^2}{12} + O(x^4) \right) F=−12−x212−x26+O(x4)F = -\frac{1}{2} - \frac{x^2}{12} - \frac{x^2}{6} + O(x^4) F=−12+x2(−112−212)+O(x4)F = -\frac{1}{2} + x^2\left(-\frac{1}{12} - \frac{2}{12}\right) + O(x^4) F=−12−3x212+O(x4)=−12−x24+O(x4)F = -\frac{1}{2} - \frac{3x^2}{12} + O(x^4) = -\frac{1}{2} - \frac{x^2}{4} + O(x^4)

Now, A=eFA = e^F: A=e−1/2−x2/4+O(x4)=e−1/2e−x2/4+O(x4)A = e^{-1/2 - x^2/4 + O(x^4)} = e^{-1/2} e^{-x^2/4 + O(x^4)} Using eu=1+u+O(u2)e^u = 1+u+O(u^2): A=e−1/2(1+(−x24+O(x4))+O(x4))A = e^{-1/2} \left( 1 + \left(-\frac{x^2}{4} + O(x^4)\right) + O(x^4) \right) A=e−1/2(1−x24+O(x4))A = e^{-1/2} \left( 1 - \frac{x^2}{4} + O(x^4) \right)

Finally, substitute AA and BB back into the limit expression: lim⁡x→0A−Bx2=lim⁡x→0e−1/2(1−x24+O(x4))−e−1/2(1−x28+O(x4))x2\lim_{x \to 0} \frac{A-B}{x^2} = \lim_{x \to 0} \frac{e^{-1/2} \left( 1 - \frac{x^2}{4} + O(x^4) \right) - e^{-1/2} \left( 1 - \frac{x^2}{8} + O(x^4) \right)}{x^2} =lim⁡x→0e−1/2(1−x24−1+x28+O(x4))x2= \lim_{x \to 0} \frac{e^{-1/2} \left( 1 - \frac{x^2}{4} - 1 + \frac{x^2}{8} + O(x^4) \right)}{x^2} =lim⁡x→0e−1/2(−2x28+x28+O(x4))x2= \lim_{x \to 0} \frac{e^{-1/2} \left( -\frac{2x^2}{8} + \frac{x^2}{8} + O(x^4) \right)}{x^2} =lim⁡x→0e−1/2(−x28+O(x4))x2= \lim_{x \to 0} \frac{e^{-1/2} \left( -\frac{x^2}{8} + O(x^4) \right)}{x^2} =lim⁡x→0e−1/2(−18+O(x2))= \lim_{x \to 0} e^{-1/2} \left( -\frac{1}{8} + O(x^2) \right) =−18e−1/2= -\frac{1}{8} e^{-1/2}

The final answer is −18e\boxed{-\frac{1}{8\sqrt{e}}}.

Solution: The limit is of the form 00\frac{0}{0}. We use Maclaurin series expansions.

  1. Expand csc⁡2x\csc^2 x: csc⁡2x=1x2(1+x23+O(x4))\csc^2 x = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4)).
  2. Expand the common exponent E=csc⁡2x(cos⁡x−1)E = \csc^2x(\cos x - 1): cos⁡x−1=−x22+x424−O(x6)\cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - O(x^6). E=1x2(1+x23+O(x4))(−x22+x424−O(x6))=−12−x28+O(x4)E = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4))(-\frac{x^2}{2} + \frac{x^4}{24} - O(x^6)) = -\frac{1}{2} - \frac{x^2}{8} + O(x^4).
  3. Expand the second term ecsc⁡2x(cos⁡x−1)=eEe^{\csc^2x(\cos x -1)} = e^E: eE=e−1/2−x2/8+O(x4)=e−1/2(1−x28+O(x4))e^E = e^{-1/2 - x^2/8 + O(x^4)} = e^{-1/2}(1 - \frac{x^2}{8} + O(x^4)).
  4. Expand the first term (cos⁡x)csc⁡2x=ecsc⁡2xln⁡(cos⁡x)(\cos x)^{\csc^2x} = e^{\csc^2x \ln(\cos x)}. Let F=csc⁡2xln⁡(cos⁡x)F = \csc^2x \ln(\cos x). ln⁡(cos⁡x)=ln⁡(1+(cos⁡x−1))=(cos⁡x−1)−(cos⁡x−1)22+O((cos⁡x−1)3)\ln(\cos x) = \ln(1 + (\cos x - 1)) = (\cos x - 1) - \frac{(\cos x - 1)^2}{2} + O((\cos x - 1)^3). ln⁡(cos⁡x)=(−x22+x424)−12(−x22)2+O(x6)=−x22−x412+O(x6)\ln(\cos x) = (-\frac{x^2}{2} + \frac{x^4}{24}) - \frac{1}{2}(-\frac{x^2}{2})^2 + O(x^6) = -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6). F=1x2(1+x23+O(x4))(−x22−x412+O(x6))=−12−x24+O(x4)F = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4))(-\frac{x^2}{2} - \frac{x^4}{12} + O(x^6)) = -\frac{1}{2} - \frac{x^2}{4} + O(x^4). So, (cos⁡x)csc⁡2x=eF=e−1/2−x2/4+O(x4)=e−1/2(1−x24+O(x4))(\cos x)^{\csc^2x} = e^F = e^{-1/2 - x^2/4 + O(x^4)} = e^{-1/2}(1 - \frac{x^2}{4} + O(x^4)).
  5. Substitute these expansions into the limit: lim⁡x→0e−1/2(1−x24+O(x4))−e−1/2(1−x28+O(x4))x2\lim_{x \to 0} \frac{e^{-1/2}(1 - \frac{x^2}{4} + O(x^4)) - e^{-1/2}(1 - \frac{x^2}{8} + O(x^4))}{x^2} =lim⁡x→0e−1/2(1−x24−1+x28+O(x4))x2= \lim_{x \to 0} \frac{e^{-1/2}(1 - \frac{x^2}{4} - 1 + \frac{x^2}{8} + O(x^4))}{x^2} =lim⁡x→0e−1/2(−x28+O(x4))x2=−18e−1/2= \lim_{x \to 0} \frac{e^{-1/2}(-\frac{x^2}{8} + O(x^4))}{x^2} = -\frac{1}{8}e^{-1/2}.

The final answer is −18e\boxed{-\frac{1}{8\sqrt{e}}}.